hdu4704之费马小定理+整数快速幂

SumTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 589 Accepted Submission(s): 292


Problem Description
Sample Input

2

Sample Output

2Hint1.  For N = 2, S(1) = S(2) = 1.2.  The input file consists of multiple test cases.  

/*分析:题目要求s1+s2+s3+...+sn;//si表示n划分i个数的n的划分的个数,如n=4,则s1=1,s2=3假设An=s1+s2+s3+...+sn;对于n可以先划分第一个数为n,n-1,n-2,...,1，则容易得出An=A0+A1+A2+A3+...+A(n-1);=>A(n+1)=A0+A1+A2+A3+...+An =>An=2^(n-1);由于n非常大,所以这里要用到费马小定理:a^(p-1)%p == 1%p == 1;//p为素数所以2^n%m == ( 2^(n%(m-1))*2^(n/(m-1)*(m-1)) )%m == (2^(n%(m-1)))%m * ((2^k)^(m-1))%m == (2^(n%(m-1)))%m;//k=n/(m-1)*/#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<iomanip>#define INF 99999999using namespace std;const int MAX=100000+10;const int mod=1000000000+7;char s[MAX];__int64 MOD(char *a,int Mod){__int64 sum=0;for(int i=0;a[i] != '\0';++i){sum=(sum*10+a[i]-'0')%Mod;}return sum;}__int64 FastPow(__int64 a,__int64 k){k=(k+mod)%mod;__int64 sum=1;while(k){if(k&1)sum=sum*a%mod;a=a*a%mod;k>>=1;}return sum;}int main(){while(scanf("%s",s)!=EOF){__int64 n=MOD(s,mod-1)-1;printf("%I64d\n",FastPow(2,n));}return 0;}